# In triangle ABC, K and L are points on BC. K is closer to B than L. AL is the bisector of angle KAC. KL times BC=BK times CL. Find angle BAL.

In $\mathrm{\Delta }ABC,$, K and L are points on BC. AL is the bisector of $\mathrm{\angle }KAC$. $KL×BC=BK×CL$. Find $\mathrm{\angle }BAL$.
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Julien Zuniga
Step 1
The identity $KL\cdot BC=BK\cdot CL$ implies that line AB is the external bisector of triangle CKA at A while AL is the interior bisector at A.
Step 2
Hence they are orthogonal and thus the angle $\mathrm{\angle }\phantom{\rule{thinmathspace}{0ex}}BAL={90}^{\circ }$.
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Hagman7v
Step 1
By the bisector theorem we have $\frac{AC}{AK}=\frac{CL}{KL}=\frac{BC}{BK}$ hence $\mu =\frac{AC}{CB}=\frac{AK}{KB}$ and K belongs to the Apollonius circle that is the locus of points P such that $\frac{PA}{PB}=\lambda$. Such a circle goes through the feet of the angle bisectors (the internal and the external) from C. If ABC is an acute-angled triangle, such a circle intersects the BC segment only at C, hence there is no solution.

On the other hand, if we allow $\stackrel{^}{BAC}$ to be greater than ${90}^{\circ }$, we have a solution at $\stackrel{^}{BAL}={90}^{\circ }$: